A Seifert framing has zero self-linking

By [A1], the ${\mathrm{A}\mathrm{C}}_{\omega }$ hypotheses in [F1], [F3], and [F6] hold. The given $F$ is a Seifert surface by [F1]. Use its inward unit conormal as the reference normal field. By [F2], this Seifert framing has relative winding integer zero.

Let $s$ be the inward unit conormal, viewed in the round-metric normal bundle. By [F8], choose a smooth collar $c:K\times [0,{\epsilon }_0)\to F$ and orient its levels by the original $K$. For sufficiently small $\delta >0$, set $J_0=c(K,\delta )$. It is embedded, lies in $\mathrm{int}F$, and is disjoint from $K$. No prescribed collar velocity is assumed: since $c$ is an embedding onto a boundary neighbourhood, ${\mathrm{\partial }}_{t}c(p,0)$ is inward transverse to $K$ in $F$. Its normal quotient is therefore $a(p)s(p)$ for a smooth function $a(p)>0$; a tangential component is allowed.

Use a normalized tubular chart $\mathrm{\Phi }$ of $K$ as in [F4] and [F9], and write ${\mathrm{\Phi }}^{-1}(c(p,t))=(b_t(p),v_t(p))$, with $v_t(p)$ in the normal fibre over $b_t(p)$. Compactness permits one small interval of $t$ on which these expressions are defined for every $p$. At $t=0$ they are $(p,0)$, and normalization identifies the vertical derivative with $a(p)s(p)$. The maps $b_t$ are $C^1$-close to the identity. In a smooth oriented circle coordinate their lifts have the form $u+r_t(u)$ with $r_t$ periodic and $\mathrm{\parallel }{r}_t^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}2$ for small $t$; thus the lifts are increasing degree-one diffeomorphisms. Hence each $b_t$ is an orientation-preserving diffeomorphism. Reparametrize the level over $x=b_t(p)$, obtaining a smooth normal section $w_t(x)=v_t(b_t^{-1}(x))$. Taylor expansion in finitely many bundle charts gives, uniformly on $K$, $w_t(x)=ta(x)s(x)+O(t^2)$. Because $a$ has positive minimum and $s$ is unit, after decreasing $\delta$ this implies $⟨w_{\delta }(x),s(x)⟩>0$ for every $x$.

Since $F$ is a compact embedded surface, it is closed and has empty interior in $S^3$; hence its complement contains a point $q$. Choose an orientation-preserving stereographic chart from $q$. It places $K\cup J_0$ and $F$ in ${\mathbb{R}}^3$. Let $F_0$ be the closure in $F$ of the complement of the open boundary collar $c(K\times [0,\delta ))$. Then $F_0$ is compact, is disjoint from $K$, and has boundary set $J_0$. Give $F_0$ its orientation inherited from $F$; at its new boundary the outward collar-normal points back toward $K$, so the induced boundary orientation is the transported orientation of $J_0$, parallel to $K$. Apply [F6] and [F7] to choose a regular oriented diagram $D$ of $K\cup J_0$, with the projection-plane orientation induced from $S^3$.

For $0\le u\le 1$, interpolate in each normal fibre by $z_u(x)=(1-\chi (u))w_{\delta }(x)+\chi (u)\delta s(x)$, where $\chi (u)=\sigma (3u-1)$ is the plateau from [F10]. Every $z_u$ has positive pairing with $s$, so it never meets the zero section. Uniformly these vectors have norm $O(\delta )$; a common disk bundle about the compact zero section lies in the tubular domain, so all interpolating graphs lie in that domain when $\delta$ is small. Each graph projects identically to $K$, hence is an embedded circle, and $\mathrm{\Phi }$ carries it to an embedded circle disjoint from $K$. With $K$ as the fixed first component this is a smooth two-component link isotopy, stationary near its parameter endpoints, from $K\cup J_0$ to $K\cup \mathrm{\Phi }(\delta s)$. The orientation on $J_0$ agrees with the graph orientation because $b_{\delta }$ preserves orientation. By [F9] the family extends to an ambient isotopy and its linking number is unchanged. The endpoint $\mathrm{\Phi }(\delta s)$ is exactly a sufficiently small normalized-chart push-off in [F3]–[F4]. Consequently the self-linking in the statement equals $\mathrm{lk}(K,J_0)$, without identifying the collar level literally with a chart graph of $\delta s$.

Apply [F5] with its first component $J$ equal to the original knot $K$, its second (boundary) component equal to $J_0$, and the oriented spanning surface $\mathrm{\Sigma }=F_0$. The curve $K$ is disjoint from $F_0$, so it is transverse to $F_0$ with empty intersection and has signed intersection number zero. Thus [F5] gives ${\mathrm{lk}}_D(K,J_0)=\mathrm{lk}(K,J_0)=0$. This computes the linking number in the same order as the self-linking convention. The Axiom of Choice is the hypothesis used by [F5]; [A1] supplies the countable-choice hypotheses of the other cited suppliers.

By [F3] and step 3.1, $\mathrm{lk}(K,J_0)$ is the self-linking of the Seifert framing; step 3.2 makes it zero. Step 1.1 gives relative framing integer zero.